\(\int F^{c (a+b x)} \sqrt {g \cos (d+e x)} \sqrt {f \sin (d+e x)} \, dx\) [138]

Optimal result

Integrand size = 34, antiderivative size = 136 \[ \int F^{c (a+b x)} \sqrt {g \cos (d+e x)} \sqrt {f \sin (d+e x)} \, dx=-\frac {F^{c (a+b x)} \sqrt {g \cos (d+e x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {-e-i b c \log (F)}{4 e},\frac {1}{4} \left (3-\frac {i b c \log (F)}{e}\right ),e^{4 i (d+e x)}\right ) \sqrt {f \sin (d+e x)}}{\sqrt {1-e^{2 i (d+e x)}} \sqrt {1+e^{2 i (d+e x)}} (i e-b c \log (F))} \] Output:

-F^(c*(b*x+a))*(g*cos(e*x+d))^(1/2)*hypergeom([-1/2, 1/4*(-e-I*b*c*ln(F))/ 
e],[3/4-1/4*I*b*c*ln(F)/e],exp(4*I*(e*x+d)))*(f*sin(e*x+d))^(1/2)/(1-exp(2 
*I*(e*x+d)))^(1/2)/(1+exp(2*I*(e*x+d)))^(1/2)/(I*e-b*c*ln(F))
 

Mathematica [A] (verified)

Time = 10.63 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.88 \[ \int F^{c (a+b x)} \sqrt {g \cos (d+e x)} \sqrt {f \sin (d+e x)} \, dx=\frac {i F^{c (a+b x)} \sqrt {g \cos (d+e x)} \, _2F_1\left (-\frac {1}{2},-\frac {1}{4}-\frac {i b c \log (F)}{4 e};\frac {3}{4}-\frac {i b c \log (F)}{4 e};e^{4 i (d+e x)}\right ) \sqrt {f \sin (d+e x)}}{\sqrt {1-e^{4 i (d+e x)}} (e+i b c \log (F))} \] Input:

Integrate[F^(c*(a + b*x))*Sqrt[g*Cos[d + e*x]]*Sqrt[f*Sin[d + e*x]],x]
 

Output:

(I*F^(c*(a + b*x))*Sqrt[g*Cos[d + e*x]]*HypergeometricPFQ[{-1/2, -1/4 - (( 
I/4)*b*c*Log[F])/e}, {3/4 - ((I/4)*b*c*Log[F])/e}, E^((4*I)*(d + e*x))]*Sq 
rt[f*Sin[d + e*x]])/(Sqrt[1 - E^((4*I)*(d + e*x))]*(e + I*b*c*Log[F]))
 

Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int F^{c (a+b x)} \sqrt {f \sin (d+e x)} \sqrt {g \cos (d+e x)} \, dx\)

\(\Big \downarrow \) 7271

\(\displaystyle \frac {\sqrt {g \cos (d+e x)} \int F^{c (a+b x)} \sqrt {\cos (d+e x)} \sqrt {f \sin (d+e x)}dx}{\sqrt {\cos (d+e x)}}\)

\(\Big \downarrow \) 7271

\(\displaystyle \frac {\sqrt {f \sin (d+e x)} \sqrt {g \cos (d+e x)} \int F^{c (a+b x)} \sqrt {\cos (d+e x)} \sqrt {\sin (d+e x)}dx}{\sqrt {\sin (d+e x)} \sqrt {\cos (d+e x)}}\)

\(\Big \downarrow \) 7292

\(\displaystyle \frac {\sqrt {f \sin (d+e x)} \sqrt {g \cos (d+e x)} \int F^{a c+b x c} \sqrt {\cos (d+e x)} \sqrt {\sin (d+e x)}dx}{\sqrt {\sin (d+e x)} \sqrt {\cos (d+e x)}}\)

\(\Big \downarrow \) 7299

\(\displaystyle \frac {\sqrt {f \sin (d+e x)} \sqrt {g \cos (d+e x)} \int F^{a c+b x c} \sqrt {\cos (d+e x)} \sqrt {\sin (d+e x)}dx}{\sqrt {\sin (d+e x)} \sqrt {\cos (d+e x)}}\)

Input:

Int[F^(c*(a + b*x))*Sqrt[g*Cos[d + e*x]]*Sqrt[f*Sin[d + e*x]],x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 7271

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*v^m)^ 
FracPart[p]/v^(m*FracPart[p]))   Int[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, 
x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&  !(Eq 
Q[v, x] && EqQ[m, 1])
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7299

Int[u_, x_] :> CannotIntegrate[u, x]
 

Maple [F]

\[\int F^{c \left (b x +a \right )} \sqrt {g \cos \left (e x +d \right )}\, \sqrt {f \sin \left (e x +d \right )}d x\]

Input:

int(F^(c*(b*x+a))*(g*cos(e*x+d))^(1/2)*(f*sin(e*x+d))^(1/2),x)
 

Output:

int(F^(c*(b*x+a))*(g*cos(e*x+d))^(1/2)*(f*sin(e*x+d))^(1/2),x)
 

Fricas [F]

\[ \int F^{c (a+b x)} \sqrt {g \cos (d+e x)} \sqrt {f \sin (d+e x)} \, dx=\int { \sqrt {g \cos \left (e x + d\right )} \sqrt {f \sin \left (e x + d\right )} F^{{\left (b x + a\right )} c} \,d x } \] Input:

integrate(F^(c*(b*x+a))*(g*cos(e*x+d))^(1/2)*(f*sin(e*x+d))^(1/2),x, algor 
ithm="fricas")
 

Output:

integral(sqrt(g*cos(e*x + d))*sqrt(f*sin(e*x + d))*F^(b*c*x + a*c), x)
 

Sympy [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \sqrt {g \cos (d+e x)} \sqrt {f \sin (d+e x)} \, dx=\text {Timed out} \] Input:

integrate(F**(c*(b*x+a))*(g*cos(e*x+d))**(1/2)*(f*sin(e*x+d))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int F^{c (a+b x)} \sqrt {g \cos (d+e x)} \sqrt {f \sin (d+e x)} \, dx=\int { \sqrt {g \cos \left (e x + d\right )} \sqrt {f \sin \left (e x + d\right )} F^{{\left (b x + a\right )} c} \,d x } \] Input:

integrate(F^(c*(b*x+a))*(g*cos(e*x+d))^(1/2)*(f*sin(e*x+d))^(1/2),x, algor 
ithm="maxima")
 

Output:

integrate(sqrt(g*cos(e*x + d))*sqrt(f*sin(e*x + d))*F^((b*x + a)*c), x)
 

Giac [F]

\[ \int F^{c (a+b x)} \sqrt {g \cos (d+e x)} \sqrt {f \sin (d+e x)} \, dx=\int { \sqrt {g \cos \left (e x + d\right )} \sqrt {f \sin \left (e x + d\right )} F^{{\left (b x + a\right )} c} \,d x } \] Input:

integrate(F^(c*(b*x+a))*(g*cos(e*x+d))^(1/2)*(f*sin(e*x+d))^(1/2),x, algor 
ithm="giac")
 

Output:

integrate(sqrt(g*cos(e*x + d))*sqrt(f*sin(e*x + d))*F^((b*x + a)*c), x)
 

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \sqrt {g \cos (d+e x)} \sqrt {f \sin (d+e x)} \, dx=\int F^{c\,\left (a+b\,x\right )}\,\sqrt {g\,\cos \left (d+e\,x\right )}\,\sqrt {f\,\sin \left (d+e\,x\right )} \,d x \] Input:

int(F^(c*(a + b*x))*(g*cos(d + e*x))^(1/2)*(f*sin(d + e*x))^(1/2),x)
 

Output:

int(F^(c*(a + b*x))*(g*cos(d + e*x))^(1/2)*(f*sin(d + e*x))^(1/2), x)
 

Reduce [F]

\[ \int F^{c (a+b x)} \sqrt {g \cos (d+e x)} \sqrt {f \sin (d+e x)} \, dx=\frac {\sqrt {g}\, f^{a c +\frac {1}{2}} \left (2 f^{b c x} \sqrt {\sin \left (e x +d \right )}\, \sqrt {\cos \left (e x +d \right )}-\left (\int \frac {f^{b c x} \sqrt {\sin \left (e x +d \right )}\, \sqrt {\cos \left (e x +d \right )}\, \cos \left (e x +d \right )}{\sin \left (e x +d \right )}d x \right ) e +\left (\int \frac {f^{b c x} \sqrt {\sin \left (e x +d \right )}\, \sqrt {\cos \left (e x +d \right )}\, \sin \left (e x +d \right )}{\cos \left (e x +d \right )}d x \right ) e \right )}{2 \,\mathrm {log}\left (f \right ) b c} \] Input:

int(F^(c*(b*x+a))*(g*cos(e*x+d))^(1/2)*(f*sin(e*x+d))^(1/2),x)
 

Output:

(sqrt(g)*f**((2*a*c + 1)/2)*(2*f**(b*c*x)*sqrt(sin(d + e*x))*sqrt(cos(d + 
e*x)) - int((f**(b*c*x)*sqrt(sin(d + e*x))*sqrt(cos(d + e*x))*cos(d + e*x) 
)/sin(d + e*x),x)*e + int((f**(b*c*x)*sqrt(sin(d + e*x))*sqrt(cos(d + e*x) 
)*sin(d + e*x))/cos(d + e*x),x)*e))/(2*log(f)*b*c)